Course: Introductory Physics Experiments 1

MINISTRY OF EDUCATION AND TRAINING

NONG LAM UNIVERSITY

SCIENCES DEPARTMENT

Physics Group

Course: Introductory Physics Experiments 1

Academic year: 2011 – 2012

CONTENTS

Unit 1: Errors theory ..................................................................................................... 1

Unit 2: Measuring the size of objects by micrometer and vernier caliper .................... 7

Unit 3: Determining the wavelength of monochromatic light .................................... 19

Unit 4: Measuring the mass density of a solid using the jar method .......................... 25

Unit 5: Measuring the mass density of a liquid using the hydrostatic method ........... 29

Unit 6: Determining the surface tension coefficent of a liquid ................................... 33

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Experiments Physics 1

Nong Lam University - Sciences Department - Physics Group 1

UNIT 1: ERRORS THEORY

When measuring a quantity we must use instruments and our senses.

Therefore, we cannot determine the exact value of the quantity to be measured. On

making different measurements of a particular quantity we often obtain different

values; it means that we have errors on this quantity.

There are different types of errors and each type of error required different

calculation. For example, a quantity to be measured, called X, after several

measurements have been made and then analyzed, the results must be wriiten as

follows:

X X X [1.1]

Where X is the quantity to be measured, X is the avarage of X, and X the

average absolute error on measuring X.

1. The average value of quantity X

It is denoted by X and given by

1

n

i

i

X

Xn

[1.2]

where n is the number of measurements made on X, and X i is the value of X

obtained in the ith measurement.

2. The average absolute error

It is denoted by X . This quantity gives the uncertainty of the value to be

obtained of X and is given by

random instrumentX X X [1.3]

where randomX is the average error due to the randomness and instrumentX is that due

to instrumentation. This quantity gives the uncertainty on measuring X.

3. The average error due to randomness

It is denoted by randomX . This type of error is due to the inherently random

nature of measuring directly a particular quantity many times. It depends on several

factors: the quantity to be measured, the accuracy of the instruments, human’s limited

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capability of reading, and external environment. As a result, each measurement of the

same quantity gives a little bit different values.

randomX is given by

1

n

i

irandom

X X

Xn

[1.4]

3.The instrumental error

It is denoted by instrumentX . This type of error is due to the lack of accuracy of

the instruments.

It is calculated as follows:

Reading the value on using the instrument. For example, on the micrometer,

the lowest division is 0.01mm, thus instrumentX 0.01mm.

Getting 1

2or 1 of value of the lowest division. As a result, for example, on the

micrometer, the lowest division is 0.01mm, thus instrumentX = 0.01mm/2 =

0.005 mm.

For electrical instruments such as ammeter, voltmeter, etc., we have by

convention

instrumentX = 1% Xmax

where Xmax is the lowest division of the instrument used.

Noting that, in these experiment we only get the value of instrumentX

instrumentX the lowest division [1.5]

Consequently, the instrument should be chosen so that the instrument’s

measuring scale is at the nearest to the value to be measured. The instrumental

error is then small, giving the smaller value of X and better accuracy of the

measurement.

Combining equations [1.2] , [1.3] , [1.4] , [1.5] we obtain the final result [1.1]

4. The average relative error

It is denoted by and given by

X

X

[1.6]

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The average relative error is used to evaluate the accuracy of the measurement.

Therefore, to evaluate the accuracy of the measurement we need to find both the

average relative error and the average absolute error.

5. Formulae to find errors indirectly

This is shown on the table below.

Function of the quantity to

be measure, X(a, b, c)

The formulae to calculate errors

The average absolute error

X

The average relative error

X

X

X = a + b + c X a b c

a b c

a b c

X = a – b X a b a b

a b

X = a.b . .X a a b b a b

a b

X = an

1( ) nX n a a

an

a

7. Writing the results and rounding the numbers

The value of the quantity to be measured is written in the form X X X ,

obeying the following rules:

(1) The value of X is written in the standardized form: .10nX a with 1

< a < 10.

(2) The value of X is normally written with a non-zero digit and in

accordance with the power value n of X .

(3) Discard the unreliable digits in X (the digit whose order is smaller

than the power value n of X ).

Example: We have raw values: X = 297.159 and X = 0.2765, now

follow the steps

(1) Standardixing X : X = 2.97159 x 102

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(2) Keeping a non-zero digit and rounding X such there is an

accordance in the power between X and X : X = 0.003 x 102

(3) Discarding the unreliable digits in X and rounding the result

numbers.

X = 0.003 x 102

Thus: X = 2.972 x 102

Finally: X = (2.972 0.003) x 102

Example:

Determine the result of measurment of density D of the solid using the

formula 1 20

3 2

m mD d

m m. The data are given

3

1 2 3 018.55 0.01 10.02 0.01 13.17 0.01 ; 1.00 0.05 / m g ; m g ; m g d g cm

Calculating D

31 2 1 20 0

3 2 3 2

18.55 10.021.00 2.7079 /

13.17 10.02

m m m mD d D d g cm

m m m m

Calculating D

Getting natural logarithm both sides of the equation for calculating D, we have

1 2 3 2 0ln ln( ) ln( ) ln D m m m m d

Differentiating this equation

3 2 01 2

1 2 3 2 0

( ) ( )( )

( ) ( )

d m m d dd m mdD

D m m m m d

After some manipulations, it reduces to

1 3 01 2 3

1 2 1 2 3 2 3 2 0

( )1 1

( )( )

m m d ddDdm dm dm

D m m m m m m m m d

Converting to D

D

:

First significant digit

X = 2 . 9 7 1 5 9 x102

significant digits

Unreliable digits

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1 3 01 2 3

1 2 1 2 3 2 3 2 0

( )1 1

( )( )

m m dDm m m

dD m m m m m m m m

Substituting the numbers, we have 0.059

D

D

We now calculate: 30.059 2.7079 0.059 0.159766 / D D g cm The result

received

32.7079 /D g cm (in standardized form)

3 30.159766 / 0.2 / D g cm g cm : keeping one significant digit

Rewriting to make sure an accordance in power: 32.7 /D g cm

Final result: 32.7 0.2 / D g cm

8. Drawing a diagram

In many experiments, we often need to represent the results by drawing a diagram

based on the collected data, depicting the relationship between two quantities of

interest.

How to draw a diagram

(1) Make a data table.

(2) Setting up the coordinates (an ordinate and an abscissa).

(3) Putting units on the two axes

(4) Choose proper scales

(5) Drawing error rectangles whose centers are coincident with the

measured values ( X and y ) and sides are 2 x and 2 y , respectively.

X

X

y

y

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(6) Drawing a smooth curve connecting those rectangles.

1X

2X

3X

3y

2y

1y

12 x

12 y

22 x

22 y

32 x

32 y

0 x

y



UNIT 2: MEASURING THE SIZE OF OBJECTS BY MICROMETER and

VERNIER CALIPER

The purpose of the experiment

Help the students know to use vernier to increase accuracy of measurement

and to adjust to Zero correction.

A/ MEASURING THE SIZE OF OBJECTS BY MICROMETER

Micrometer is an instrument measuring the size of objects with high accuracy

(from 0.01mm to 0.02 mm).

I. Theoretical backgrounds

1. Construction of outside micrometer

Frame: The frame is made of cast iron on which the whole assembly is

attached.

Barrel/Sleeve: Main scale is graduated on it with 0.5 mm distance.

Thimble: Divisions are marked on this cylindrical portion. Knurling is made

on it for better gripping.

Spindle: Spindle is one of the contacting surface during measurement. It

moves to and fro as thimble rotated.

Anvil: It is the another contacting surface which is fixed in Frame.

Lock nut: This nut is used to lock the movement of the spindle.

Ratchet: Ratchet is used to rotate and apply uniform pressure during

measurement.

Figure 2.1



A micrometer comprising a main body holding an anvil at one end portion

thereof and a spindle moving to and away from the anvil at the other end portion

thereof through an inner sleeve, an outer sleeve covering the outside of the inner

sleeve and fixed thereon, and a thimble unitedly connected to the spindle on the

outside of the outer sleeve in a manner that the thimble rotates, said outer sleeve

being provided with graduations and numerals on the outer circumferential surface

along the axial direction, and the thimble being provided with graduations and

numerals on the outer circumference thereof.

2. Graduation of Outside micrometer

In metric micrometer the pitch of the spindle thread is 0.5 mm. Thereby in one

rotation of the thimble, the spindle advances by 0.5 mm. On the barrel (sleeve) a 25

mm long datum line is marked. This line is further graduated to millimeters and half

millimeters. The graduations are numbered as 0, 5, 10, 15, 20 and 25 mm.

The circumference of the bevel edge of the thimble is graduated into 50

divisions and marked 0-5-10-15...45-50 in a clockwise direction. The distance moved

by the spindle during one rotation of the thimble is 0.5 mm.

3. Reading of Outside micrometer

Step 1: First note the minimum range of the outside micrometer. While measuring

with a 50 to 75 mm micrometer, note it as 50 mm.

Figure 2.2



Step 2: The least count of a Metric micrometer is 0.01 mm. It is the ratio of value of 1

main scale division and total graduations on thimble. i.e. 0.5/50.

Step 3: Then read the barrel (sleeve) graduations. Read the value of the visible lines

on the left of the thimble edge.

Step 4: Next, read the thimble graduations. Multiply these graduations with least

count.

Step 5: Add all above readings together. The sum of these readings is the actual

reading of the object.

Example of Reading – 1

Figure 2.3



Example of Reading - 2

Example of Reading – 3

Figure 2.5

Figure 2.4



Example of reading – 4

4. Zero correction. Always check whether full closure of the jaws actually

gives a zero reading. Special wrenches are available to set the zero reading exactly.

Alternatively, the zero reading may be treated as a correction value to be added to

(or subtracted from) all readings made with the instrument. This value is called a

"zero correction."

II. Instruments

(1) Micrometers

(2) Some copper wires

(3) Some metallic balls

III. Practice

1. Adjusting "zero correction" of the micrometer

2. Measuring the diameter of a copper wire: A wire is measured five times

at five different positions on the wire.

3. Measuring the diameter, d, of a ball: measure five times the diameter of

a ball at five different positions on the ball, then calculate the ball’s

volume V and its error V .

31

6V d và 3

dV

d

IV. Reporting (see the reporting form of unit 2A)

Figure 2.6



THE REPORTING FORM OF UNIT 2A

1. Use the micrometer to measure diameters d1, d2 of two steel balls and those of six

copper wires. Then establish the result tables as follows:

Table 2.1: The diameters of two steel balls

Measurement d1 d1(random) d2 d2(random)

1

2

3

4

5

The average

value

Table 2.2: The diameters of 2 copper wires

Measurement D1 D1(random) D2 D2(random)

1

2

3

4

5

The average value

2. Calculating the volumes of two steel balls ( 1ball V , ball 2V )

3. Calculating the corresponding errors

a. 1D , 2D ,

b. 1

ball V ,

2

ball V , Vball 1, Vball 2

where ball 1 = ball 1

ball 1

V

V

, ball 2 = ball 2

ball 2

V

V

Writing the results

D1 = 1D 1D D2 = 2D 2D

Vball 1 = 1 ball 1ball V V Vball 2 = ball 2ball 2V V



4. Show usages of the vernier in a micrometer. Give some examples. In order to obtain

an accuracy of 0.005 mm for the micrometer, how does its contruction have to be changed?

Describe this micrometer’s construction then.

B/ MEASURING THE SIZE OF OBJECTS BY VERNIER CALIPER

I. Theoretical backgrounds

1. Construction of the vernier caliper

1 - Outside jaws: used to take external measures of objects

2 - Inside jaws: used to take internal measures of objects

3 - Depth probe: used to measure the depth of objects

4 - Main scale (cm)

5 - Main scale (inch)

6 - Vernier (cm)

7 - Vernier (inch)

8 - Retainer: used to block movable part

The vernier caliper consists of two components

The fixed component: rule T is divided into mm, connecting jaw A and jaw

A’. A and A’ lie on a straight line.

The sliding component: Depth probe C is attached to edge B and edge B’ of

the vernier caliper. It can move on rule T. B and B’ lie on another straight line

that is parallel to the line connecting A and A’

A B

A’ B’

01 02

Figure 2.7



2. Reading a vernier Caliper

A vernier caliper is used to precisely measure dimensions to within 0.01 mm. It’s

important that you learn to read it properly – a small error in measurement may have

a great effect on subsequent calculations.

A picture of a typical vernier caliper appears below. Note that it can be used to

measure inner and outer dimensions of objects, as well as the depth of a hole. You

will usually be using the larger “jaws” to measure outer dimensions.

Make sure that you read the correct scale. The fixed portion of the scale is

typically marked in increments of 1.0 mm; the sliding scale is marked in 0.01 mm

increments.

3. Measurements are taken as follows:

Step 1: Close the sliding jaw so that it fits snugly on the object to be measured. If the

object is circular or spherical, make sure you’re measuring at the widest point.

Figure 2.8



Step 2: Remove the caliper from the object without changing the position of the sliding

jaw.

SStteepp 33:: FFiirrsstt,, rreeaadd tthhee ““00”” ppoossiittiioonn ooff tthhee vveerrnniieerr ssccaallee ((sslliiddiinngg ssccaallee)) oonn tthhee mmaaiinn ssccaallee

((ffiixxeedd ssccaallee)) ttoo ggeett aa rroouugghh rreeaaddiinngg..

SStteepp 44:: NNeexxtt,, llooookk aalloonngg tthhee vveerrnniieerr ssccaallee uunnttiill oonnee ooff tthhee vveerrnniieerr ddiivviissiioonn ccooiinncciiddeess wwiitthh

tthhee mmaaiinn ssccaallee,, ttoo ggeett aa rreeaaddiinngg oonn vveerrnniieerr ssccaallee.. Multiply these graduations with least

count.

Figure 2.9

Figure 2.11

Figure 2.12



Step 5: Obtain aaccccuurraattee rreeaaddiinngg Add all above readings together. The sum of these

readings is the actual reading of the object to be measured.

II. Instruments

(1) The vernier caliper

(2) The quill cylinder

(3) The rectangular cylinder

III. Procedure

Practice using the vernier caliper by measuring several objects. Check your

measurements with your lab partner; if there is disagreement, check the vernier scale

again. Be sure to ask if you’re still unsure as to the correct procedure!

1. Measuring capacity of vessel

2

4V d h

d: The inner diameter

h: Depth of the object (using the Depth probe)

2. Measuring volume of rectangular block

V =𝒂 × 𝒃 × 𝒄

Measuring five times the sides a, b, c of the block at different positions.

IV. Reporting: (see the reporting form of unit 2B)

Figure 2.13

C

Figure 2.14



THE REPORTING FORM OF UNIT 2B

1. Use the vernier caliper to measure

Diameter d and height h2 of a cylinder which has a base (cylinder);

Sides a, b, c of a rectangular block

Depths h1, h2, h3, h4,h5, h6.

Then, put the results obtained into the below tables:

Table 3.1: A cylinder which has a base

Measurement d drandom h hrandom

1

2

3

4

5

The average

value

Table 3.2: Sides a, b, c of retangular block

Measurement a arandom b brandom c crandom

1

2

3

4

5

The average

value

2. Calculating the volumes of the three samples mentioned above

;cylinder rectangular V V

3. Calculating the corresponding errors:



, d h , , a b c

4. Calculating the errors

, cylinder rectangular

cylinder rectangular , ; V VV V

5. Writing the results

Vcylinder = cylindercylinder

V V

Vretangular = rectangularrectangular

V V

Show usages of the vernier caliper. Give some examples. In order to obtain an

accuracy of 0.01 mm for the vernier caliper, how does its contruction have to be changed?

Describe ernier caliper’s construction then.



UNIT 3: DETERMINING THE WAVELENGTH OF

MONOCHROMATIC LIGHT

I. The purposes of the experiment

Observing the interference of white light with Young’s double-slit experiment.

Determining the wavelength of monochromatic light with Young’s double-slit

experiment.

This experiment is to make student familiar to using instruments to produce

the interferent phenomenon.

II. Theoretical backgrounds

YOUNG ’ S DOUBLE-SLIT EXPERIMENT

Young used a monocharomatic light source (primary source, S) and projected

the light onto two very narrow slits (S1 & S2), as shown in Figure 3.1. The light from

the source will then diffract through the slits, and the patterns can be projected onto a

screen. Since there is only one source of light, the set of two waves which emanate

from the slits will be in phase with each other.

As a result, these two slits, denoted as S1 and S2 , serve as a pair of coherent

light sources. The light waves from S1 and S2 produce on a viewing screen a visible

pattern of bright and dark parallel bands called fringes, as shown in Figure 3.2. When

the light from S1 and that from S2 both arrive at a point on the screen such that

Figure 3.1: Schematic diagram of Young ’ s double-slit experiment. Two

slits behave as coherent sources of light waves that produce an interference pattern on the viewing screen (drawing not to scale).



constructive interference occurs at that location, a bright fringe appears. When the

light from the two slits combines destructively at any location on the screen, a dark

fringe results.

We can describe Young ’ s experiment quantitatively with the help of Figure

3.3. The viewing screen is located a perpendicular distance L from the double- slitted

barrier. S1 and S2 are separated by a distance d, and the source is monochromatic. To

reach any arbitrary point P, a wave from the lower slit travels farther than a wave

from the upper slit by a distance 𝑑𝑠𝑖𝑛𝜃. This distance is called the path difference 𝛿

(lowercase Greek delta).

If we assume that two rays, S1P and S2P, are parallel, which is approximately true

because L is much greater than d, then 𝛿 is given by

Figure 3.2: A typical pattern from a two-slit experiment of

interference.

Figure 3.3: Geometric construction for describing Young’s double-slit

experiment (not to scale).



𝜹 = 𝑺𝟐𝑷 − 𝑺𝟏𝑷 = 𝒓𝟐 − 𝒓𝟏 = 𝒅𝒔𝒊𝒏𝜽 [3.1]

where d = S1S2 is the distances between the two coherent light sources (i.e., the two

slits).

If 𝛿 is either zero or some integer multiple of the wavelength, then the two

waves are in phase at point P and constructive interference results. Therefore, the

condition for bright fringes, or constructive interference, at point P is

𝜹 = 𝒓𝟐 − 𝒓𝟏 = 𝒏𝝀 [3.2]

where n = 0, ±1, ±2 , ….

The number n in equation [3.2] is called the order number. The central bright fringe

at 𝜃 = 0 (n = 0) is called the zeroth-order maximum. The first maximum on either

side, where n = ±1, is called the first-order maximum, and so forth.

When 𝛿 is an odd multiple of n/2, the two waves arriving at point P are 180°

out of phase and give rise to destructive interference. Therefore, the condition for

dark fringes, or destructive interference, at point P is

𝜹 = 𝒓𝟐 − 𝒓𝟏 = (𝒏 +𝟏

𝟐)𝝀 [3.3]

where n = 0, ±1, ±2, ....

It is useful to obtain expressions for the positions of the bright and dark fringes

measured vertically from O to P. In addition to our assumption that L >> d, we

assume that d >> λ . These can be valid assumptions because in practice L is often of

the order of 1 m, d a fraction of a millimeter, and λ a fraction of a micrometer for

visible light. Under these conditions, 𝜃 is small; thus, we can use the approximation

𝑠𝑖𝑛𝜃 ≈ 𝑡𝑎𝑛𝜃. Then, from triangle OPQ in Figure 3.3, we see that

𝒚 = 𝑶𝑷 = 𝑳𝒕𝒂𝒏𝜽 ≈ 𝑳𝒔𝒊𝒏𝜽 [3.4]

From equations [3.1], [3.2] and [3.4], we can prove that the positions of the bright

fringes measured from O are given by the expression

𝒚𝒃𝒓𝒊𝒈𝒉𝒕 = 𝒏𝝀𝑳

𝒅 [3.5]

Similarly, using equations [3.1], [3.3] and [3.4], we find that the dark fringes are

located at

𝒚𝒅𝒂𝒓𝒌 = (𝒏 + 𝟏/𝟐)𝝀𝑳

𝒅 [3.6]



Using the formular [3.5] and [3.6], we determine the distance between two

adjacent bright fringe [𝑦𝑏𝑟𝑖𝑔ℎ𝑡(𝑛+1) − 𝑦𝑏𝑟𝑖𝑔ℎ𝑡(𝑛)] or distance between two adjacent

dark fringe [𝑦𝑑𝑎𝑟𝑘 (𝑛+1) − 𝑦𝑑𝑎𝑟𝑘 𝑛 ]. This resulf is given by

𝒊 = 𝒚𝒃𝒓𝒊𝒈𝒉𝒕(𝒏+𝟏) − 𝒚𝒃𝒓𝒊𝒈𝒉𝒕(𝒏) = 𝒚𝒅𝒂𝒓𝒌(𝒏+𝟏) − 𝒚𝒅𝒂𝒓𝒌 𝒏 =𝝀𝑳

𝒅 [3.7]

As we demonstrate in the following example, Young’s double-slit experiment

provides a method for measuring the wavelength of light. If we have m adjacent

bright fringes, we will have (m – 1) of value i in [3.7], therefore, we determine the

value of λ. In fact, Young used this technique to do just that. Additionally, the

experiment gave the wave model of light a great deal of credibility. It was

inconceivable that particles of light coming through the slits could cancel each other

in a way that would explain the dark fringes. As a result, the light interference show

that light is of wave nature.

III. Instruments

(1) A solid-state lamp (laze) 1÷1,5mW

(2) A plate has two slits. These slits are parallel to each other, separated by

a distance d = 0.15mm (Young’s double-slit).

(3) A viewing screen.

(4) Some of bases.

(5) A measuring tape (tape-line) into milimeter.

(6) Transformer

(𝟏) (𝟐) (𝟑)

(𝟒) (𝟔)

Figure 3.4: The experiment arrangement.



IV. Procedure

Step 1: Attch the solid-state lamp and the Young-slit to the bracket. (?)

Step 2: Connect the lamp to the alternating source of 220V and adjust the Young-slit

so that the light from the lamp can shine through the Young-slit.

Step 3: Put the viewing screen on the bracket. The viewing screen and Young-slit are

separated by a distance 1m (L1 = 1m). Use the measuring tape to get value l1, the

distance of five adjacent bright fringes or five adjacent dark fringes.

Step 4: Calcultate 𝒊𝟏 =𝒍𝟏

𝟒=

𝝀𝑳𝟏

𝒅 and after that find the value of 𝜆 with 𝝀 =

𝒅𝒊𝟏

𝑳𝟏

Repeat these above steps five times with the other value of L and determine the

corresponsding values of 𝝀.



THE REPORTING FORM OF UNIT 3

(1) Question: Write the equation of λ in terms 𝑑, 𝑙,𝐿 and the corresponding

significance of these quantities.

(2) Experiment’s result: Finish the below data sheet and calculate the average

values and corresponding errors.

Given: 𝑑 = 0.150 𝑚𝑚 ± 0.005𝑚𝑚 is the distance between S1 and S2.

measurment L(mm) Δ𝐿 𝑙(𝑚𝑚) Δ𝑙 𝜆 =

𝑑𝑙

4𝐿

1

2

3

4

5

Average

value

The final result 𝝀 = 𝝀 ± ∆𝝀



Unit 4: MEASURING MASS DENSITY OF A SOLID USING THE JAR

METHOD

I. The purpose of the experiment

This experiment is to make student familiar to the method of scaling

repeatedly.


1. Mass density

The mass density D of a substance of uniform composition is its mass

per unit volume:

objectmD

V [4.1]

mobject : The mass of the object (kg)

V: The volume of the object (m3)

2. Method of measuring

If the object has a simple shape such as a sphere, a cube, ect., first we can use

vernier caliper or micrometer to determine its volume and use a scale to find

the mass of the object.

The object has a complicated shape such we can use find the object’s volume

by comparing its mass with the mass of an amount of water, which has the

same volume as that of the object. This method is called jar method

0waterm

DV

[4.2]

waterm : The mass of the water

D0: The mass density of the water

From equations [4.1] and [4.2], we have

0

object

water

mD D

m [4.3]



III. Instruments

(1) A balance and a set of weights

(2) A glass jar

(3) Water

(4) The solid object whose density is to be measured

(5) Wipers (using to clean the glass jar)

IV. Procedure

With a balance, use the method of repeated scalings to measure the mass of

the object.

Step 1: Make the balance stable on a horizontal plane by using a plumb -line

(adjusting screws at the balance’s stand).

Step 2: Repeat weighings in an order as follows:

a. The first weighing (Figure 4.1)

Put intermediate object B on the left scale of the balance and put a jar full of

water and several weights on the right scale of the balance. Adjust the amount of

weights to make the total mass on the right scale equal to the mass on the left scale.

mB = mwater + jar + m1 [4.4]

mB: The mass of object B

m1: The total mass of the weights on the

right scale after adjusting

mwater+jar: The mass of the jar full of water

b. The second weighing (Figure 3.2)

Keep the same object B on the left

scale, put the solid object of interest on the

right scale, and take out some weights

until the two masses on both scale are

equal

mB = mwater+jar + mobject + m2 [4.5]

Figure 4.2

B

m2

object

Bell jar has a

capillary tube

Figure 4.1

B



mobject: The mass of the solid object of interest .

m2: The total mass of those weights still on the right scale.

c. The third weighing (Figure 4.3)


scale, put the solid object of interest into

the jar still full of water. Some water will

be displaced out of the jar via a capillary,

whose volume is called Vwater/out. The

volume of the object of interest, Vobject, is

equal to Vwater/out. Clean gently the jar until its outside is dry then adjust the weights

on the right scale until the two masses on both scales are equal

mB = m’water + jar + mobject + m3 [4.6]

m’water+jar: The mass of the jar plus the mass of water still in the jar.

From equations [4.4] và [4.5], we have mobject = m1 - m2

From equations [4.4] và [4.5]: we have mwater/out = m3 – m2

mwater/out: The mass of the water which is displaced out of the jar.

Putting those results in equation [4.3], we have

1 20

3 2

m mD D

m m

[4.7]

Step3: Repeat step 2, each weighting is carried five times, and record the

foundings.

V. Reporting: (see the reporting form of unit 4)

Figure 4.3

B

m3




1. Describe the method used to measure the mass density of a solid body which

has a complicated shaped. Give the procedure to establish the equation

calculating the mass density of this body.

2. Weigh the mass m1, m2, m3. Each weighting is carried out five times.

3. Calculating the mass densities D1, D2, D3, D4, D5 then from those densities

find the average density D .


+ 321 ,, mmm

+ ,D D , D

D

D

or

D

D

D

Writing the results D =D D

Given

+ The accuracy of the balance is 10 mg.

+ D0 = (996.0 0.1)kg/m3

Table 4.1: The mass m1, m2, m3.

Measurement m1 m2 m3 1( )randomm 2( )randomm 3( )randomm

1

2

3

4

5

The average

value

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UNIT 5: MEASURING THE MASS DENSITY OF A LIQUID USING THE

HYDROSTATIC METHOD

I. The purpose of the experiment

This experiment is to make student familiar to the method of scaling

repeatedly.


The mass density of the fluid is ratio of the mass of the liquid and the mass of

the water having the same volume

The formula for calculating the mass density of a liquid

liquid

water

m

m [5.1]

liquidm The mass of an amount of the liquid of interest

waterm The mass of an amount of water, having the same volume as that of

the liquid.

To determine fluidm , waterm , we compare the Archimedes forces (buoyant

forces) acting on an object which is submerged in turn in the liquid and the

water - hydrostatic method

III. Instruments

(1) A balance and a set of weights

(2) An intermediate object B

(3) A pycnometer

(4) An amount of the fluid of interest

(5) Water



IV. Procedure

Step 1: Make the balance stable on a horizontal plane by using a plumb -line

(adjusting screws at the balance’s stand)

Step 2: Repeat weighings in an order as follows:

(a) The first weighing (Figure 5.1):

Put intermediate object B on the

left scale of the balance and several

weights and a hanging object on the right

scale of the balance. Adjusting weighing

to the amount of weights to make the

total mass on the right scale equal to the

mass on the left scale.

1B hanging objectm m m [5.2]

mB: The mass of the object B

mhanging-object : The mass of the hanging object on the right scale

m1: The total mass of the weights on the right scale after adjusting.

(b) The second weighing (Figure 5.2)


scale, put the hanging object on the right

scale into the liquid of interest, and some

weights until the two masses on both scale

are equal

2B hanging object liquidm m m m

[5.3]

Lm g : The weight of the liquid which is

displaced, it is equal to Archimedes force

(buoyant force)

Where

mhanging-object: The mass of the solid object of interest.

m2: The total mass of those weights still on the right scale.

Figure 5.1

B m1

Object

Liquid

Figure 5.2

B m2



(c) The third weighing (Figure 5.3):


scale, put the hanging object on the right

scale into water, and put in or take out

some weights until the two masses on

both scales are equal

3B hanging object waterm m m m

[5.4]

Caution: We must wipe the hanging object

off the residue liquid before putting it into the water.

From equations [5.2] và [5.3], we have

2 1liquidm m m [5.5]

From equations [5.2] và [5.4], we have

3 1 waterm m m [5.6]

Put the results from [5.5], [5.6] into [5.1]. We obtain

2 1

3 1

m m

m m [5.7]

Step3: Repeat step 2, each weighting is carried five times, and record the

foundings.


Figure 5.3

B m3

water




1. Describe the method used to measure the mass density of a liquid. Give the

procedure to establish the equation calculating the mass density of a liquid by

using the hydrostatic method.

2. Weigh the mass m1, m2, m3. Each weighting is carried out five times.

3. Calculating the mass density 1, 2, 3, 4, 5 then from those densities find the

average density .


+ 321 ,, mmm

+ , where

Writing the results =

Given, the accuracy of the balance is 10 mg.

Table 5.1: The mass m1, m2, m3

Measurement m1 m2 m3 1( )randomm 2( )randomm 3 ( )n randomm

1

2

3

4

5

The average

value



UNIT 6: DETERMINING THE SURFACE TENSION COEFFICIENT OF A

LIQUID

I. The purposes of the experiment

Examine capillary phenomenon in a thin glass tube containing a liquid.

Use results of the examination to determine the surface tension coefficient

of the liquid.


Phenomenon: Dip a thin glass tube with a rather small inside diameter into

water such that it is at right angle with the water surface. We will see that the level of

the liquid inside the tube is different from that of the liquid outside the tube.

The capillary phenomenon is the one in which the surface of a liquid where it

contacts a solid is elevated or depressed due to the force by which a liquid is drawn

along a very narrow tube. Such a tube is called a capillary tube.

Actually, the existence of surface tension gives rise to a hydrostatic pressure

difference, ΔP, between two fluids separated by a curved interface.

2P

R

Where is the surface tension coefficient in the interface, and R is the radius

of the interfacial curvature.

M N

h

R

Figure 6.1



We examine the case in which the liquid surface level inside the tube is

elevated. Consider two point M and N at the same level, as shown in Figure 6.1

The pressure at N

PN = P0 [6.1]

Where P0 is the atmospheric pressure.

The pressure at M

0MP P gh P [6.2]

gh : The static pressure due to the water column of height h; ρ is the

mass density of liquid; g is acceleration due to gravity.

Because M and N are of the same level, we have

PM = PN [6.3]

From equations [6.1], [6.2], [6.3], we have

Ph

g

[6.4]

In this case the capillary tube is a thin cylinder of inside diameter r, the curved

interface is concave. The curved interface has the form a spherical top whose radius

is R

rR

cos

[6.5]

Where is the contact angle and R is negative because the the curved

interface is concave.

ΔP is given by

2P

R

[6.6]

Where σ is the surface tension in the interface.

- From equations [6.5], [6.6], we have

2h

R g

[6.7]

- From equations [6.5], [6.7] we have

2 4cos cosh

R g d g

[6.8]

http://www.answers.com/topic/density

http://www.answers.com/topic/acceleration

http://www.answers.com/topic/gravity

http://www.answers.com/topic/contact-angle



Where d is the inside diameter of the tube.

Equation [6.8] expresses the Jurin’s rule which is used to find the height of

the liquid column inside the capillary tube.

In our lab, the liquid used is alcohol. Because alcohol wets fully the glass, the

contact angle θ is zero. Thus, ffrom equation [6.8], we obtain

4

hd g

[6.9]

III. Instruments

(1) A capillary tube

(2) A disk containing alcohol

(3) A vertical clamp

(4) An inverted observing tool

IV. Procedure

Step 1: Dip the capillary tube into the alcohol, move it up and down several

times, then clamp it vertically.

Step 2: Turn a screw at the inverted observing tool such that one of its division

lines is coincident to the alcohol level of the liquid inside the tube. Record height h1

of this level.

capillary tube

h1 h2

0

Figure 6.2

http://www.answers.com/topic/contact-angle



Turn the screw again such that one of another division line is coincident to the

alcohol level outside the tube. Record the height h2 of this level.

Then we have

h = h1 – h2 [6.10]

Carry step 1 and step 2 five times and change the position of the tube on the

alcohol surface.

Use equation [6.9] to calculate the surface tension coefficient σ of the liquid.





1. What phenomenon occurs on the surface of the liquid ? Describe the

experiment and explain this phenomenon.

2. Describe how to measure the surface tension coefficient of a liquid through

the capillary phenomenon.

3. Use an observing tool to measure height h1, h2. This is carried out five times.

4. Calculating

+ ,h , where is the surface tension coefficient of the liquid.

+ ,,, 21 hh ,

or

Writing the results:

Given

+ The wetting angle of the liquid is 30o. ( 030 )

+ 3(996.0 0.1) /kg m

+ d = 1.00 0.01 mm

+ g = 29.86 0.02 /m s

Table : The heights, h1 and h2, are measured by observing tool

Measurment h1 h2 h 1( )randomh

2( )randomh

1

2

3

4

5

The average

value

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Course: Introductory Physics Experiments 1

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